1.

5x+2y=2k2(k+1)x+ky=(3k+4)

Answer» {tex}5x + 2y = 2k{/tex}{tex}2(k + 1)x + ky = (3k + 4){/tex}These are of the form{tex}a_1x + b_1y + c_1\xa0= 0 , a_2x + b_2y + c_2\xa0= 0{/tex}where,{tex}a_1= 5\xa0,b_1= 2, c_1\xa0= -2k,{/tex}{tex}a_2= 2(k +1)\xa0,b_2= k\xa0,c_2\xa0= -(3k + 4){/tex}For infinitely many solutions, we must\xa0have{tex}\\frac { a _ { 1 } } { a _ { 2 } } = \\frac { b _ { 1 } } { b _ { 2 } } = \\frac { c _ { 1 } } { c _ { 2 } }{/tex}This holds only when{tex}\\frac { 5 } { 2 ( k + 1 ) } = \\frac { 2 } { k } = \\frac { - 2 k } { - ( 3 k + 4 ) }{/tex}{tex}\\Rightarrow \\frac { 5 } { 2 ( k + 1 ) } = \\frac { 2 } { k } = \\frac { 2 k } { ( 3 k + 4 ) }{/tex}Now, the following cases arises:Case 1:{tex}\\frac { 5 } { 2 ( k + 1 ) } = \\frac { 2 } { k }{/tex}[Taking I and II]{tex}\\Rightarrow 5 k = 4 ( k + 1 ) \\Rightarrow 5 k = 4 k + 4{/tex}k = 4Case 2:{tex}\\frac { 2 } { k } = \\frac { 2 k } { ( 3 k + 4 ) }{/tex}[Taking II and III]{tex}\\Rightarrow{/tex}2(3k +\xa04) = 2k2\xa0{tex}\\Rightarrow{/tex}6k + 8 = 2k2{tex}\\Rightarrow{/tex}{tex}2k^2\xa0-\xa06k + 8 =\xa00{/tex}{tex}\\Rightarrow{/tex}{tex}2(k^2\xa0- 3k + 4)=\xa00{/tex}{tex}\\Rightarrow{/tex}{tex}k^2\xa0- 3k - 4 = 0{/tex}{tex}\\Rightarrow{/tex}{tex}k^2\xa0- 4k + k - 4 = 0{/tex}{tex}\\Rightarrow{/tex}k(k - 4) + 1(k - 4) = 0{tex}\\Rightarrow{/tex}(k - 4)(k +\xa01) = 0(k - 4) = 0 or k + 1 = 0{tex}\\Rightarrow{/tex}\xa0k = 4 or k = -1Case 3:{tex}\\frac { 5 } { 2 ( k + 1 ) } = \\frac { 2 k } { ( 3 k + 4 ) }{/tex}[Taking I and II]{tex}\\Rightarrow 15 k + 20 = 4 k ^ { 2 } = 4 k{/tex}{tex}\\Rightarrow{/tex}4k2\xa0+ 4k - 15k - 20 = 0{tex}\\Rightarrow{/tex}\xa04k2\xa0- 11k - 20 = 0{tex}\\Rightarrow{/tex}4k2\xa0- 16k + 5k - 20 =0{tex}\\Rightarrow{/tex}4k(k - 4) + 5(k - 4) = 0{tex}\\Rightarrow{/tex}(k - 4)(4k + 5) = 0{tex}\\Rightarrow k = 4 \\text { or } k = \\frac { - 5 } { 4 }{/tex}Thus, k = 4, is the common value of which there are infinitely many solutions.


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